Differentiation (Part-38) | S N De

 

$\,3.\,$ Express $\,0.\overline{4}\,\,$ as an infinite geometric series and hence prove that $\,\,(0.444\cdots)^{1/2}=0.666\cdots$

Sol. $\,\,0.\overline{4}\\=0.4444\cdots \infty\\=0.4+0.04+0.004+0.0004+\cdots \infty\\=\frac{4}{10}+\frac{4}{10^2}+\frac{4}{10^3}+\frac{4}{10^4}+\cdots\infty \rightarrow(1)$

Clearly, $\,(1)\,$ represents an infinite geometric series with the first term $=\frac{4}{10},\,$ common ratio $(r)=\frac{4/10^2}{4/10}=\frac{1}{10};\,\,$ where $\,-1<r<1\,$ and so the finite value of sum exists for the given infinite geometric series.

Hence, from $\,(1),\,$ we get

$\,0.\overline{4}=\frac{4}{10} \cdot \frac{1}{1-\frac{1}{10}}\\~~~~~~=\frac{4}{10} \cdot \frac{1}{9/10}\\~~~~~~=\frac{4}{10} \times \frac{10}{9}\\~~~~~~=\frac 49$

Now, $\,\,(0.444\cdots)^{1/2}\\=(\frac 49)^{1/2}\\=\left[\frac{2^2}{3^2}\right]^{1/2}\\=\left(\frac 23\right)^{2\times \frac 12}\\=\frac 23=0.666\cdots \,\,\text{(proved)}$

$\,4.\,$ The sum of an infinite geometric series is $\,\frac 13\,$ and its first term is $\,\frac 14;\,$ find the series.

Sol. Let common ratio of the infinite geometric series is $\,r\,\,(-1<r<1)$ so that 

$\,~~~~S=\frac{a}{1-r}\,\,[*] \\ \Rightarrow \frac 13=\frac{1/4}{1-r}\\ \Rightarrow \frac 13=\frac{1}{4(1-r)}\\ \Rightarrow 4-4r=3 \\ \Rightarrow r=\frac 14.$

Note[*]: Here the first term $\,(a)=\frac 14,$ sum $(S)=\frac 13.$

Hence, the infinite geometric series is given by :

$=a+ar+ar^2+\cdots \infty\\=\frac 14+\frac 14 \cdot \frac 14+\frac 14 \cdot \left(\frac 14\right)^2+\cdots \infty\\=\frac 14+\frac{1}{16}+\frac{1}{64}+\cdots \infty\,\,\text{(ans.)}$ 

$\,5.\,$ The sum of an infinite geometric series is $\,6\,$ and the sum of its first two terms is $\,\frac 92;$ find its common ratio.

Sol. Let us consider the infinite geometric series $\,a,ar,ar^2,ar^3,\cdots ;$ where $\,a\,$ is the first term and $\,r\,$ being the common ratio.

Now, $\,S=a+ar+ar^2+\cdots\infty\\~~~=\frac{a}{1-r}\,\,[\text{where}\,\,-1<r<1]\\ \Rightarrow 6=\frac{a}{1-r}\\ \Rightarrow a=6(1-r)\rightarrow(1)$

Again, $\,a+ar=\frac 92\\ \Rightarrow a(1+r)=\frac 92\\ \Rightarrow 6(1-r)(1+r)=\frac 92\,\,[\text{By (1)}]\\ \Rightarrow 6(1-r^2)=\frac 92 \\ \Rightarrow 1-r^2=\frac{9}{12} =\frac 34 \\ \Rightarrow  r^2=1-\frac 34 \\ \Rightarrow  r^2=\frac 14=\left(\frac 12\right)^2 \\ \Rightarrow r=\pm\frac 12\,\,\text{(ans)}$

$\,6.\,$ In a G.P., the sum of infinite terms is $\,15;$ the sum of squares of these terms is $\,45.$ Find the G.P.

Sol.  Let us consider the infinite geometric series $\,a,ar,ar^2,ar^3,\cdots ;$ where $\,a\,$ is the first term and $\,r\,$ being the common ratio.

Then, the sum of this series $:S=\frac{a}{1-r}=15\rightarrow(1) \\ \Rightarrow a=15(1-r)\rightarrow(2)$

Again, the squares of those terms are : $\,a^2,a^2r^2,a^3r^3,\cdots$ and in this series the first term is $\,a^2\,$and the common ratio is $\,r^2.$

Now, since the sum of squares of these terms is $\,45,\,$ so 

$\,\frac{a^2}{1-r^2}=45 \\ \Rightarrow \frac{a}{1-r} \times \frac{a}{1+r}=45 \\ \Rightarrow 15 \times \frac{a}{1+r}=45 \\ \Rightarrow \frac{a}{1+r}=\frac{45}{15}=3  \\ \Rightarrow \frac{15(1-r)}{1+r}=3\,\,[\text{By (2)}]  \\ \Rightarrow 15(1-r)=3(1+r) \\ \Rightarrow 5(1-r)=1+r \\ \Rightarrow 5-1= r+5r  \\ \Rightarrow 6r=4  \\ \Rightarrow r=\frac 46=\frac 23.$

So, by $\,(2),\,$ we get , $\,a=15(1-\frac 23)=15 \cdot \frac 13=5.$

Hence, the required infinite geometric series is :

$\,~~~a,ar,ar^2,ar^3,\cdots\\=5, 5 \cdot \frac 23, 5 \cdot (\frac 23)^2, 5 \cdot (\frac 23)^3,\cdots\\=5,\frac{10}{3},\frac{20}{9},\frac{40}{27},\cdots$

$\,7.\,$ Two infinte geometric series start from the same number. If the common ratios are $\,0.85\,$ and $\,0.55\,$ respectively, show that the sum of the first series is three times that of the second.

Sol.  Let us consider the infinite geometric series $\,a,ar,ar^2,ar^3,\cdots ;$ where $\,a\,$ is the first term and $\,r\,$ being the common ratio.

Let $\,A=a+ar+ar^2+\cdots=\frac{a}{1-0.85}\,\,[\because r=0.85]$

For the second geometric series, let us consider the infinite geometric series $\,a,ab,ab^2,ab^3,\cdots ;$ where $\,a\,$ is the first term and $\,b\,$ being the common ratio.

Let $\,B=a+ab+ab^2\cdots=\frac{a}{1-0.15}\,[\because b=0.15]$

Hence, $\,\frac AB=\frac{a/0.15}{a/0.85}=\frac{a}{0.15}\times \frac{0.85}{a} \\ \Rightarrow A=3 \times B\,\,\text{(showed)}$

$\,8.\,$ The distance passed over by a certain pendulam bob in succeeding swings form the G.P. $\,(16,12,9,\cdots)\,$cm respectively. Calculate the total distance traversed by the bob before it comes to rest.

Sol. Clearly, $\,(16,12,9,\cdots)\,$ represents a infinite geometric series with the first term $(a)=16,\,\,$ and common ratio $(r)=\frac{12}{16}=\frac 34.$

So, $\,a+ar+ar^2+\cdots\\=16+12+9+\cdots\\=\frac{a}{1-r}\\=\frac{16}{1-\frac 34}\\=\frac{16}{\frac 14}\\=16\times 4\\=64.$

Hence, the total distance traversed by the bob is $\,64\,$ cm before it comes to rest.